Theorem: all 4s shall be false

Theorem: The first one who got the number that is multiple of 4 (i.e. n % 4 == 0) will lost, otherwise he/she will win.

Proof:

the base case: when n = 4, as suggested by the hint from the problem, no matter which number that that first player, the second player would always be able to pick the remaining number.

For 1* 4 < n < 2 * 4, (n = 5, 6, 7), the first player can reduce the initial number into 4 accordingly, which will leave the death number 4 to the second player. i.e. The numbers 5, 6, 7 are winning numbers for any player who got it first.

Now to the beginning of the next cycle, n = 8, no matter which number that the first player picks, it would always leave the winning numbers (5, 6, 7) to the second player. Therefore, 8 % 4 == 0, again is a death number.

Following the second case, for numbers between (2*4 = 8) and (3*4=12), which are 9, 10, 11, are winning numbers for the first player again, because the first player can always reduce the number into the death number 8.

Following the above theorem and proof, the solution could not be simpler:

when you take 2, n = 6;
and your opponent will definitely take 2, because he or she is also very clever and have optimal strategies as it has been mentioned in the problem.
so now n = 4 and it's your turn...sorry for that ; )

If you are familiar with math induction, this problem can be proved by natural induction.

Base case: n = 4, we can show that for taking 1, 2, or 3, you will always lose.

Induction case: suppose for n = 4K, you will lose. For 4 (K + 1), you have 4K + 4, which is 4 more than 4K. For all the possible choice (1, 2, 3), your enemy can take 3, 2, 1 so you must take 4K, which is losing.

The other case it is possible to win, not that you will win. The problem is "can"win Nim.
So to clarify, I'd say that if you show that you will never win in the case where n % 4 == 0, then you show that you will "not never" win in the other cases.